/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Fairbanks Temperature The normal... [FREE SOLUTION] | 91Ó°ÊÓ

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Fairbanks Temperature The normal mean temperature at Fairbanks, Alaska, is modeled as $$ f(x)=37 \sin (0.0172 x-1.737)+25^{\circ} \mathrm{F} $$ where \(x\) is the number of days since the last day of the previous year. (Source: B. Lando and C. Lando, "Is the Curve of Temperature Variation a Sine Curve?" The Mathematics Teacher, vol.7, no. \(6,\) September \(1977,\) p. 535\()\) a. Write the rate-of-change function for the normal mean temperature at Fairbanks, Alaska. b. How rapidly was the temperature changing 180 days since the last day of the previous year? Interpret the result.

Short Answer

Expert verified
The rate of temperature change at 180 days is 0.1329°F/day, indicating the temperature is increasing.

Step by step solution

01

Understand the Problem

We need to find the rate of change of the function \( f(x) = 37 \sin(0.0172x - 1.737) + 25 \) with respect to \( x \). This requires calculating the derivative \( f'(x) \). After finding \( f'(x) \), we evaluate it at \( x = 180 \) to interpret how the temperature is changing on that day.
02

Differentiate the Function

The function \( f(x) \) is \( f(x) = 37 \sin(0.0172x - 1.737) + 25 \). The derivative of a constant is 0, and we differentiate the sine function using the chain rule. The derivative is \( f'(x) = 37 \cdot \cos(0.0172x - 1.737) \cdot 0.0172 \). Simplifying gives \( f'(x) = 0.6364 \cos(0.0172x - 1.737) \).
03

Evaluate the Derivative at x = 180

Substitute \( x = 180 \) into \( f'(x) = 0.6364 \cos(0.0172x - 1.737) \). Calculate \( 0.0172 \times 180 - 1.737 = 1.359 \). Thus, \( f'(180) = 0.6364 \cos(1.359) \). Use a calculator to evaluate \( \cos(1.359) \approx 0.2088 \). Hence, \( f'(180) = 0.6364 \times 0.2088 \approx 0.1329 \) degrees per day.
04

Interpret the Result

The positive value of \( 0.1329 \) degrees per day means that the temperature at Fairbanks, Alaska, is increasing at a rate of approximately 0.1329°F per day, 180 days after the last day of the previous year.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate of Change
The rate of change is an important concept in calculus, representing how a quantity changes with respect to another variable. In this exercise, we are interested in how the temperature in Fairbanks, Alaska changes over time. To find this, we calculate the derivative of the temperature function with respect to time, measured in days from the last day of the previous year.
The derivative, often denoted as \( f'(x) \), gives the instantaneous rate at which the temperature is changing at any given day \( x \). This helps in predicting whether the temperature is rising or falling.
  • If the derivative is positive, the temperature is increasing.
  • If it's negative, the temperature is decreasing.
  • If it's zero, the temperature is stable.
This analysis allows us to make informed predictions about weather patterns and is especially useful in climate studies.
Temperature Modeling
Temperature modeling using mathematical functions is a common technique to represent natural phenomena. In this problem, the sine function is used to simulate the variation in temperature in Fairbanks, Alaska throughout the year. The model can provide useful information on expected weather patterns, seasonal changes, and anomalies.
The equation given is \( f(x) = 37 \sin(0.0172x - 1.737) + 25 \), where \( x \) is the number of days from the previous year. This function captures the cyclical nature of temperature change, typical of many climates.
  • The amplitude \( 37 \) indicates the range of temperature fluctuation above or below the mean.
  • The phase shift \(-1.737\) adjusts for the correct starting point in the year.
  • The vertical shift \(+25\) indicates the average temperature.
Understanding and using such a model helps in strategic planning, such as preparing for seasonal temperature shifts.
Sine Function
The sine function is key to modeling periodic phenomena because of its wavelike properties. When applied to temperature modeling, as in Fairbanks' scenario, it provides a natural way to approximate the ups and downs of temperature over time.
The function \( \sin(\theta) \) oscillates between \(-1\) and \(1\), providing a smooth, continuous wave-like pattern. Here are some features of the sine function in this context:
  • The coefficient before the sine term, \( 37 \), affects the amplitude, i.e., the extent of temperature variation.
  • The argument \( (0.0172x - 1.737) \) modifies the period and phase shift, affecting how quickly and where these variations occur.
Utilizing sine functions allows us to predict periodic behaviors, such as seasonal changes in temperatures, which are known for their regular patterns.
Chain Rule
The chain rule is a fundamental tool in calculus for finding the derivative of composite functions. It is particularly useful when dealing with functions like the temperature model we're exploring.
In this exercise, we need to differentiate \( \sin(0.0172x - 1.737) \), which is a composite function involving a sine function and a linear function \((0.0172x - 1.737)\). The chain rule states that if you have a function \( y = f(g(x)) \), then the derivative \( y' \) equals \( f'(g(x)) \cdot g'(x) \).
Applying the chain rule helps us find that the derivative of the given function is \( f'(x) = 37 \cdot \cos(0.0172x - 1.737) \cdot 0.0172 \). This expression shows how we combine the derivatives of both the outer function, \( \cos \), and the inner function, \( (0.0172x - 1.737) \), to derive the temperature rate of change.
Mastering the chain rule is vital for correctly handling complex functions composed of multiple layers, as often encountered in advanced calculus problems.

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